an investigation of dual stator winding induction machines

main idea of the Butterworth method is to place the poles evenly on a half circle of radius
ω 0 in the left s-plane, with its center at origin. The four poles are given as:
5π
7π
9π
11π
j
j
j
j
8
8
8
8
1 = ω0
e , p2 = ω0
⋅ e , p3 = ω0
⋅ e , p4 = ω0
⋅ e
p ⋅
The first two poles are used to construct the Butterworth equation in complex form,
which is given as:
2
2
( − p )( p − p ) = p + ω ( 1. 3066 − j1.
3066)
p − j ⋅ω
p (10.81)
1
2
0
The observer characteristic equation is compared with Butterworth equation to find
the observer gains,
rsiL
D
i
ri
= ω
0
rsir
D
ri
i
= − j ⋅ω
rriL
+
D
i
si
L
+
D
ri
( 1.
3066 − j1.
3066)
rri
+
D
L
+ jωei
D
2
0
i
mi
i
i
mi
( K + j ⋅ K ) − j ⋅ ( 2ω
− ω ) − ( K + j ⋅ K )
11
si ri
( K + j ⋅ K ) − j ( ω − ω )
11
12
ei
ri
( K + j ⋅ K ) − j(
ω − ω ) ( K + j ⋅ K ) − ω ( ω − ω )
21
12
22
r L
D
i
ei
ei
ri
L
D
i
ri
376
ri
11
L
D
rriL
− jωei
D
i
i
si
12
0
21
ei
ei
22
ri
(10.82)
(10.83)
Both the imaginary and real parts of (10.82-10.83) need to be equal, which are
expressed as:
rsi
L
D
L
D
r
ri
i
r
i
ri
rri
Lsi
Lri
L
+ +
K
D D D
i
i
mi ( K11
) − ( 21 ) = 1. 3066 ⋅ω
0
L
mi
( K12
) − ( 2ω − ω ) − ( K 22 ) = −1.
3066 ⋅ω
0
r
ei
ri
D
i
i
mi
ri
( K ) − ( K ) + ( ω − ω ) ( K ) − ω ( ω − ω ) = 0
si ri ri
+ 11 ei
22 ei ri
12 ei ei ri
Di
Di
Di
Di
(10.84)
(10.85)
L
L
ω (10.86)